15.2 Single - Factor (One - Way) Analysis of Variance

```Adjusted Exponential
Smoothing
AFt +1 = Ft +1 + Tt +1
where
T = an exponentially smoothed trend factor
Tt +1 = (Ft +1 - Ft) + (1 - ) Tt
where
Tt = the last period trend factor
= a smoothing constant for trend
Smoothing Example
Smoothing Example
PERIOD
MONTH
DEMAND
1
2
3
4
5
6
7
8
9
10
11
12
Jan
Feb
Mar
Apr
May
Jun
Jul
Aug
Sep
Oct
Nov
Dec
37
40
41
37
45
50
43
47
56
52
55
54
Smoothing Example
PERIOD
MONTH
DEMAND
T3
= (F3 - F2) + (1 - ) T2
= (0.30)(38.5 - 37.0) + (0.70)(0)
1
2
3
4
5
6
7
8
9
10
11
12
Jan
Feb
Mar
Apr
May
Jun
Jul
Aug
Sep
Oct
Nov
Dec
37
40
41
37
45
50
43
47
56
52
55
54
= 0.45
AF3 = F3 + T3 = 38.5 + 0.45
= 38.95
T13 = (F13 - F12) + (1 - ) T12
= (0.30)(53.61 - 53.21) + (0.70)(1.77)
= 1.36
AF13 = F13 + T13 = 53.61 + 1.36 = 54.96
Smoothing Example
PERIOD
MONTH
DEMAND
FORECAST
Ft +1
1
2
3
4
5
6
7
8
9
10
11
12
13
Jan
Feb
Mar
Apr
May
Jun
Jul
Aug
Sep
Oct
Nov
Dec
Jan
37
40
41
37
45
50
43
47
56
52
55
54
–
37.00
37.00
38.50
39.75
38.37
38.37
45.84
44.42
45.71
50.85
51.42
53.21
53.61
TREND
Tt +1
FORECAST AFt +1
–
0.00
0.45
0.69
0.07
0.07
1.97
0.95
1.05
2.28
1.76
1.77
1.36
–
37.00
38.95
40.44
38.44
38.44
47.82
45.37
46.76
58.13
53.19
54.98
54.96
Smoothing Forecasts
70 –
60 –
Demand
50 –
40 –
30 –
20 –
10 –
0–
|
1
|
2
|
3
|
4
|
5
|
|
6
7
Period
|
8
|
9
|
10
|
11
|
12
|
13
Smoothing Forecasts
70 –
60 –
Actual
Demand
50 –
40 –
Forecast ( = 0.50)
30 –
20 –
10 –
0–
|
1
|
2
|
3
|
4
|
5
|
|
6
7
Period
|
8
|
9
|
10
|
11
|
12
|
13
Smoothing Forecasts
70 –
60 –
Actual
Demand
50 –
40 –
Forecast ( = 0.50)
30 –
20 –
10 –
0–
|
1
|
2
|
3
|
4
|
5
|
|
6
7
Period
|
8
|
9
|
10
|
11
|
12
|
13
Linear Trend Line
y = a + bx
where
a
b
x
y
=
=
=
=
intercept (at period 0)
slope of the line
the time period
forecast for demand for period x
Linear Trend Line
y = axy
+ bx
- nxy
b = x2 - nx2
where
a
b
x
y
= intercept
0)
a = (at
y - period
bx
= slope of the line
= where
the time period
n =for
number
of periods
= forecast
demand
for period x
x
x =
= mean of the x values
n
y
y = n = mean of the y values
Least Squares Example
x(PERIOD)
y(DEMAND)
1
2
3
4
5
6
7
8
9
10
11
12
73
40
41
37
45
50
43
47
56
52
55
54
78
557
Least Squares Example
x(PERIOD)
y(DEMAND)
xy
x2
1
2
3
4
5
6
7
8
9
10
11
12
73
40
41
37
45
50
43
47
56
52
55
54
37
80
123
148
225
300
301
376
504
520
605
648
1
4
9
16
25
36
49
64
81
100
121
144
78
557
3867
650
78
x = Example
= 6.5
Least Squares
12
x(PERIOD)
y(DEMAND)
1
2
3
4
5
6
7
8
9
10
11
12
73
40
41
37
45
50
43
47
56
52
55
54
78
557
557 x2
= 46.42
12
37
xy 1- nxy
80
4
b =
x2 9- nx2
123
148
386716- (12)(6.5)(46.42)
225 =
25
650 - 12(6.5)2
300
36
xyy =
301
376
a
504
520
605
648
3867
=
=
=
=
49
1.7264
y - bx81
100- (1.72)(6.5)
46.42
121
35.2
144
650
78
x = Example
= 6.5
Least Squares
12
x(PERIOD)
1
2
3
4
5
6
7
8
9
10
11
12
78
Linear trend line
557 x2
y(DEMAND)
xy
y
=
= 46.42
y = 35.2
+
1.72x
12
73
37
xy 1- nxy
40
80
4
b =
x2 9- nx2
41
123
37
45
50
43
47
56
52
55
54
148
225
300
301
376
a
504
520
605
648
557
3867
386716- (12)(6.5)(46.42)
25
=
650 - 12(6.5)2
36
=
=
=
=
49
1.7264
y - bx81
100- (1.72)(6.5)
46.42
121
35.2
144
650
78
x = Example
= 6.5
Least Squares
12
x(PERIOD)
1
2
3
4
5
6
7
8
9
10
11
12
78
Linear trend line
557 x2
y(DEMAND)
xy
y
=
= 46.42
y = 35.2
+
1.72x
12
73
37
1
xy
- nxy
Forecast
for
period
13
40
80
4
b =
x2 9- nx2
41
123
y148
= 35.2 + 1.72(13)
37
386716- (12)(6.5)(46.42)
45
225
y = =57.56 25
units
650 - 12(6.5)2
50
300
36
43
47
56
52
55
54
301
376
a
504
520
605
648
557
3867
=
=
=
=
49
1.7264
y - bx81
100- (1.72)(6.5)
46.42
121
35.2
144
650
Linear Trend Line
70 –
60 –
Demand
50 –
40 –
30 –
20 –
10 –
0–
|
1
|
2
|
3
|
4
|
5
|
|
6
7
Period
|
8
|
9
|
10
|
11
|
12
|
13
Linear Trend Line
70 –
60 –
Actual
Demand
50 –
40 –
30 –
20 –
10 –
0–
|
1
|
2
|
3
|
4
|
5
|
|
6
7
Period
|
8
|
9
|
10
|
11
|
12
|
13
Linear Trend Line
70 –
60 –
Actual
Demand
50 –
40 –
Linear trend line
30 –
20 –
10 –
0–
|
1
|
2
|
3
|
4
|
5
|
|
6
7
Period
|
8
|
9
|
10
|
11
|
12
|
13
 Repetitive increase/
decrease in demand
 Use seasonal factor
 Repetitive increase/
decrease in demand
 Use seasonal factor
Di
Seasonal factor = Si =
D
YEAR
1999
2000
2001
Total
DEMAND (1000’S PER QUARTER)
1
2
3
4
Total
12.6
14.1
15.3
42.0
8.6
10.3
10.6
29.5
6.3
7.5
8.1
21.9
17.5
18.2
19.6
55.3
45.0
50.1
53.6
148.7
YEAR
1999
2000
2001
Total
DEMAND (1000’S PER QUARTER)
1
2
3
4
Total
12.6
14.1
15.3
42.0
8.6
10.3
10.6
29.5
6.3
7.5
8.1
21.9
17.5
18.2
19.6
55.3
45.0
50.1
53.6
148.7
D1
42.0
S1 =
=
= 0.28
D 148.7
D3
21.9
S3 =
=
= 0.15
D 148.7
D2
29.5
S2 =
=
= 0.20
D 148.7
D4
55.3
S4 =
=
= 0.37
D 148.7
YEAR
DEMAND (1000’S PER QUARTER)
1
2
3
4
Total
1999
2000
2001
Total
12.6
14.1
15.3
42.0
8.6
10.3
10.6
29.5
6.3
7.5
8.1
21.9
17.5
18.2
19.6
55.3
Si
0.28
0.20
0.15
0.37
45.0
50.1
53.6
148.7
YEAR
DEMAND (1000’S PER QUARTER)
1
2
3
4
Total
1999
2000
2001
Total
12.6
14.1
15.3
42.0
8.6
10.3
10.6
29.5
6.3
7.5
8.1
21.9
17.5
18.2
19.6
55.3
Si
0.28
0.20
0.15
0.37
45.0
50.1
53.6
148.7
For 2002
y = 40.97 + 4.30x
= 40.97 + 4.30(4)
= 58.17
YEAR
DEMAND (1000’S PER QUARTER)
1
2
3
4
Total
1999
2000
2001
Total
12.6
14.1
15.3
42.0
8.6
10.3
10.6
29.5
6.3
7.5
8.1
21.9
17.5
18.2
19.6
55.3
Si
0.28
0.20
0.15
0.37
45.0
50.1
53.6
148.7
For 2002
y = 40.97 + 4.30x
= 40.97 + 4.30(4)
= 58.17
SF1 = (S1) (F5)
= (0.28)(58.17) = 16.28
SF3 = (S3) (F5)
= (0.15)(58.17) = 8.73
SF2 = (S2) (F5)
= (0.20)(58.17) = 11.63
SF4 = (S4) (F5)
= (0.37)(58.17) = 21.53
Forecast Accuracy
 Error = Actual - Forecast
 Find a method which minimizes error
 Mean Absolute
 Mean Absolute
Percent Deviation (MAPD)
 Cumulative Error (E)
Mean Absolute
 Dt - Ft 
n
where
t = the period number
Dt = demand in period t
Ft = the forecast for period t
n = the total number of periods
 = the absolute value
PERIOD
1
2
3
4
5
6
7
8
9
10
11
12
DEMAND, Dt
Ft ( =0.3)
37
40
41
37
45
50
43
47
56
52
55
54
37.00
37.00
37.90
38.83
38.28
40.29
43.20
43.14
44.30
47.81
49.06
50.84
557
PERIOD
1
2
3
4
5
6
7
8
9
10
11
12
DEMAND, Dt
Ft ( =0.3)
(Dt - Ft)
|Dt - Ft|
37
40
41
37
45
50
43
47
56
52
55
54
37.00
37.00
37.90
38.83
38.28
40.29
43.20
43.14
44.30
47.81
49.06
50.84
–
3.00
3.10
-1.83
6.72
9.69
-0.20
3.86
11.70
4.19
5.94
3.15
–
3.00
3.10
1.83
6.72
9.69
0.20
3.86
11.70
4.19
5.94
3.15
49.31
53.39
557
PERIOD
1
2
3
4
5
6
7
8
9
10
11
12
DEMAND, Dt
37
40
41
37
45
50
43
47
56
52
55
54
557
=
=
=
Ft ( =0.3)
(Dt - Ft)
|Dt - Ft|
37.00
–
37.00
3.00
37.90
3.10
 D
F

t
t -1.83
38.83
n
38.28
6.72
40.29
9.69
53.39
43.20
-0.20
11
43.14
3.86
44.30
11.70
47.81
4.19
4.85
49.06
5.94
50.84
3.15
–
3.00
3.10
1.83
6.72
9.69
0.20
3.86
11.70
4.19
5.94
3.15
49.31
53.39
Other Accuracy Measures
 Mean absolute percent deviation (MAPD)
|Dt - Ft|
MAPD =
Dt
 Cumulative error
E = et
 Average error
et
E= n
Comparison of
Forecasts
FORECAST
MAPD
E
(E)
Exponential smoothing (= 0.30)
Exponential smoothing (= 0.50)
(= 0.50, = 0.30)
Linear trend line
4.85
4.04
3.81
9.6%
8.5%
8.1%
49.31
33.21
21.14
4.48
3.02
1.92
2.29
4.9%
–
–
Forecast Control
 Reasons for out-of-control forecasts
 Change in trend
 Appearance of cycle
 Weather changes
 Promotions
 Competition
 Politics
Tracking Signal
 Compute each period
 Compare to control limits
 Forecast is in control if within limits
(Dt - Ft)
E
Tracking signal =
=
Use control limits of +/- 2 to +/- 5 MAD
Tracking Signal Values
PERIOD
DEMAND
Dt
FORECAST,
Ft
1
2
3
4
5
6
7
8
9
10
11
12
37
40
41
37
45
50
43
47
56
52
55
54
37.00
37.00
37.90
38.83
38.28
40.29
43.20
43.14
44.30
47.81
49.06
50.84
ERROR
Dt - Ft
–
3.00
3.10
-1.83
6.72
9.69
-0.20
3.86
11.70
4.19
5.94
3.15
E =
(Dt - Ft)
–
3.00
6.10
4.27
10.99
20.68
20.48
24.34
36.04
40.23
46.17
49.32
–
3.00
3.05
2.64
3.66
4.87
4.09
4.06
5.01
4.92
5.02
4.85
Tracking Signal Values
PERIOD
DEMAND
Dt
1
2
3
4
5
6
7
8
9
10
11
12
37
40
41
37
45
50
43
47
56
52
55
54
FORECAST,
Ft
ERROR
Dt - Ft
E =
(Dt - Ft)
37.00
–
–
37.00
3.00
3.00
37.90
3.10
6.10
38.83
-1.83
4.27
38.28
6.72 for period
10.99 3
Tracking
signal
40.29
9.69
20.68
43.20
-0.20
6.10 20.48
43.14
TS3 = 3.86 =24.34
2.00
3.05
44.30
11.70
36.04
47.81
4.19
40.23
49.06
5.94
46.17
50.84
3.15
49.32
–
3.00
3.05
2.64
3.66
4.87
4.09
4.06
5.01
4.92
5.02
4.85
Tracking Signal Values
PERIOD
DEMAND
Dt
FORECAST,
Ft
1
2
3
4
5
6
7
8
9
10
11
12
37
40
41
37
45
50
43
47
56
52
55
54
37.00
37.00
37.90
38.83
38.28
40.29
43.20
43.14
44.30
47.81
49.06
50.84
ERROR
Dt - Ft
–
3.00
3.10
-1.83
6.72
9.69
-0.20
3.86
11.70
4.19
5.94
3.15
E =
(Dt - Ft)
–
3.00
6.10
4.27
10.99
20.68
20.48
24.34
36.04
40.23
46.17
49.32
–
3.00
3.05
2.64
3.66
4.87
4.09
4.06
5.01
4.92
5.02
4.85
TRACKING
SIGNAL
–
1.00
2.00
1.62
3.00
4.25
5.01
6.00
7.19
8.18
9.20
10.17
Tracking Signal
Plot
3 –
2 –
1 –
0 –
-1 –
-2 –
-3 –
|
0
|
1
|
2
|
3
|
4
|
5
|
6
Period
|
7
|
8
|
9
|
10
|
11
|
12
Tracking Signal
Plot
3 –
2 –
Exponential smoothing ( = 0.30)
1 –
0 –
-1 –
-2 –
-3 –
|
0
|
1
|
2
|
3
|
4
|
5
|
6
Period
|
7
|
8
|
9
|
10
|
11
|
12
Tracking Signal
Plot
3 –
2 –
Exponential smoothing ( = 0.30)
1 –
0 –
-1 –
-2 –
Linear trend line
-3 –
|
0
|
1
|
2
|
3
|
4
|
5
|
6
Period
|
7
|
8
|
9
|
10
|
11
|
12
Statistical Control Charts
=
(Dt - Ft)2
n-1
 Using  we can calculate statistical
control limits for the forecast error
 Control limits are typically set at  3
Statistical Control
Charts
18.39 –
12.24 –
Errors
6.12 –
0–
-6.12 –
-12.24 –
-18.39 –
|
0
|
1
|
2
|
3
|
4
|
5
|
6
Period
|
7
|
8
|
9
|
10
|
11
|
12
Statistical Control
Charts
18.39 –
UCL = +3
12.24 –
Errors
6.12 –
0–
-6.12 –
-12.24 –
-18.39 –
|
0
LCL = -3
|
1
|
2
|
3
|
4
|
5
|
6
Period
|
7
|
8
|
9
|
10
|
11
|
12
Causal Modeling with
Linear Regression
 Study relationship between two
or more variables
 Dependent variable y depends
on independent variable x
y = a + bx
Linear Regression Formulas
a = y-bx
xy - nxy
b =
x2 - nx2
where
a = intercept (at period 0)
b = slope of the line
x
x =
= mean of the x data
n
y
y = n = mean of the y data
Linear Regression
Example
x
(WINS)
y
(ATTENDANCE)
xy
x2
4
6
6
8
6
7
5
7
36.3
40.1
41.2
53.0
44.0
45.6
39.0
47.5
145.2
240.6
247.2
424.0
264.0
319.2
195.0
332.5
16
36
36
64
36
49
25
49
49
346.7
2167.7
311
Linear Regression
Example
49
x
(WINS)
4
6
6
8
6
7
5
7
49
x=
= 6.125
8y
(ATTENDANCE)
xy
346.9
y=
= 43.36
8
36.3
145.2
40.1 - nxy2
240.6
xy
b = 41.2
247.2
x2 - nx2
53.0
424.0
(2,167.7)
- (8)(6.125)(43.36)
44.0
264.0
=
2
45.6(311) - (8)(6.125)
319.2
39.0
195.0
= 4.06
47.5
332.5
a = y346.7
- bx
2167.7
= 43.36 - (4.06)(6.125)
= 18.46
x2
16
36
36
64
36
49
25
49
311
Linear Regression
Example
49
x
(WINS)
4
6
6
8
6
7
5
7
49
x=
= 6.125
8y
(ATTENDANCE)
xy
x2
346.9
Regression
equation
y=
= 43.36
8
36.3
145.2+ 4.06x
16
y = 18.46
40.1 - nxy2
240.6
36
xy
Attendance forecast
for367 wins
b = 41.2
247.2
x2 - nx2
y = 18.46
53.0
424.0+ 4.06(7)
64
(2,167.7)
- (8)(6.125)(43.36)
44.0
264.0 or 46,880
36
= 46.88,
=
2
45.6(311) - (8)(6.125)
319.2
49
39.0
195.0
25
= 4.06
47.5
332.5
49
a = y346.7
- bx
2167.7
= 43.36 - (4.06)(6.125)
= 18.46
311
Linear Regression
Line
60,000 –
Attendance, y
50,000 –
40,000 –
30,000 –
20,000 –
10,000 –
|
0
|
1
|
2
|
3
|
4
|
5
Wins, x
|
6
|
7
|
8
|
9
|
10
Linear Regression
Line
60,000 –
Attendance, y
50,000 –
40,000 –
30,000 –
Linear regression line,
y = 18.46 + 4.06x
20,000 –
10,000 –
|
0
|
1
|
2
|
3
|
4
|
5
Wins, x
|
6
|
7
|
8
|
9
|
10
Correlation and Coefficient
of Determination
 Correlation, r
 Measure of strength of relationship
 Varies between -1.00 and +1.00
 Coefficient of determination, r2
 Percentage of variation in dependent
variable resulting from changes in
the independent variable
Computing Correlation
r=
n xy -  x y
[n x2 - ( x)2] [n y2 - ( y)2]
(8)(2,167.7) - (49)(346.9)
r=
[(8)(311) - (49)2] [(8)(15,224.7) - (346.9)2]
r = 0.947
Coefficient of determination
r2 = (0.947)2 = 0.897
Multiple Regression
Study the relationship
of demand to two or more
independent variables
y =  0 +  1x 1 +  2x 2 … +  kx k
where
0 = the intercept
1, … , k = parameters for the
independent variables
x1, … , xk = independent variables
```